Question #215697

Find the envelope of the family of the curves

1) y=3px-p^3

Where p=parameter

2) y=mx-2am-am^3

Where m=parameter


1
Expert's answer
2021-07-12T12:20:57-0400

1) The parametric equations of the envelope are defined by the system of equations


y=3pxp3y=3px-p^30=3x3p20=3x-3p^2

p=±xp=\pm\sqrt{x}

y=3xxxx=2xxy=3x\sqrt{x}-x\sqrt{x}=2x\sqrt{x}

Or


y=3xx+xx=2xxy=-3x\sqrt{x}+x\sqrt{x}=-2x\sqrt{x}

2) The parametric equations of the envelope are defined by the system of equations


y=mx2amam3y=mx-2am-am^30=x2a3am20=x-2a-3am^2

m=±x2a3am=\pm\sqrt{\dfrac{x-2a}{3a}}

y=xx2a3a2ax2a3ax2a3x2a3ay=x\sqrt{\dfrac{x-2a}{3a}}-2a\sqrt{\dfrac{x-2a}{3a}}-\dfrac{x-2a}{3}\sqrt{\dfrac{x-2a}{3a}}

=3x6ax+2a3x2a3a=\dfrac{3x-6a-x+2a}{3}\sqrt{\dfrac{x-2a}{3a}}

=2(x2a)3/233a=\dfrac{2(x-2a)^{3/2}}{3\sqrt{3a}}

Or


y=xx2a3a+2ax2a3a+x2a3x2a3ay=-x\sqrt{\dfrac{x-2a}{3a}}+2a\sqrt{\dfrac{x-2a}{3a}}+\dfrac{x-2a}{3}\sqrt{\dfrac{x-2a}{3a}}=3x+6a+x2a3x2a3a=\dfrac{-3x+6a+x-2a}{3}\sqrt{\dfrac{x-2a}{3a}}

=2(x2a)3/233a=-\dfrac{2(x-2a)^{3/2}}{3\sqrt{3a}}


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